Cauchy Principal Value Integral

Date: 09/17/2002 at 23:33:36
From: John Harter
Subject: Improper Integral Theory
Why isn't an integral from negative infinity to positive infinity
defined as the limit as t (or any other variable) approaches infinity
of the integral from -t to t? In other words, why must we split an
integral into two halves and evaluate each as a one-sided improper
integral (as the standard definition of such an integral states)?
For example, it would be intuitively pleasing for the integral of x^3
(or another odd function like x / (x^2+1), x, etc.) from negative
infinity to positive infinity to be zero. Yet this integral diverges
under the present definition because each side diverges! The fact that
the sides are equal and opposite doesn't even come into play.
This does not make sense to me. If someone has some insight into why
mathematicians haved defined integrals from negative infinity to
positive infinity this way, I would greatly appreciate it if they
would share their knowledge with me. I know there must be some good
reason for it. Thank you.

Date: 09/18/2002 at 03:13:50
From: Doctor Pete
Subject: Re: Improper Integral Theory
Hi John,
Your question raises a very good point, one that is often inadequately
discussed in undergraduate calculus courses.
However, it is important to realize that what may make intuitive sense
is not necessarily what is consistent with other mathematical
concepts. In calculus, we must be careful in deciding what it really
means to compute an improper integral, and more generally, the related
notions of convergence and indeterminate forms.
But should you feel like the type of improper integral you are
describing does not have its own meaning or significance in
mathematics, do not worry. You have just described what is known as
the Cauchy principal value integral. There are essentially two types:
the first is the doubly infinite improper integral, e.g.,
G = Integrate[F[x], {x,-Infinity,Infinity}]
for which the Cauchy principal value is defined as
PV(G) = Limit[Integrate[F[x], {x,-R,R}], R -> Infinity].
The second is where F has a discontinuity at some point c on the
interval [a,b]. Then the improper integral
H = Integrate[F[x], {x,a,b}]
has Cauchy principal value
PV(H) = Limit[Integrate[F[x], {x,a,c-e}]
+ Integrate[F[x], {x,c+e,b}], e -> 0].
In both cases, the idea is to consider the improper integrals as
limits that are symmetrically computed about the point of
discontinuity (in the first case, the point of discontinuity is at
infinity).
In light of this, you are probably wondering why the improper integral
itself is not defined to be the Cauchy principal value. After all, why
talk about the two things as if they were separate? The simple answer
is that they are indeed not the same thing at all. The key difference
here is that the improper integral must satisfy a more strict
criterion for convergence than the CPV, as required under the
definition of integral as a Riemann sum. We need to distinguish these
two things because not every function has a CPV integral, and those
that do, do not necessarily have a convergent improper integral. So a
distinction is made between the two because there are contexts in
which it is inappropriate to give an integral a CPV, and others in
which it is necessary.
We have now arrived at the heart of your question: Why not give a
function like F[x] = x^3 the Cauchy principal value when integrating
over the entire real line? Well, the function itself increases
(decreases) without bound as x tends to infinity (-infinity), and
therefore talking about the value of the improper integral is like
saying Infinity - Infinity = 0, which is not necessarily the case as
a discussion of indeterminate forms and L'Hopital's Rule would reveal.
In fact, we don't even need to consider a function that is unbounded
at infinity, only that the improper integral over a half-infinite
subinterval is unbounded. Simply put, the symmetry of the function is
unimportant in light of the fact that we are dealing with an
indeterminate form.
Another way to look at it: What if you shifted the point at which you
partition the real line? You would not want the value of the improper
integral to depend on where you choose this point. And while it may
be the case that there is an "obvious" choice for functions like
F[x] = x^3, what about arbitrary polynomials of odd degree, which are
not necessarily symmetric about the origin or any other point? It is
pretty obvious that if you shift this point, the value of the improper
integral will change. You could always choose to split the real line
at the origin, but this merely amounts to computing the Cauchy
principal value. Hence the CPV of doubly infinite integrals lacks a
very important quality: independence from the underlying coordinate
system. Therefore, we want to maintain a sufficiently strict
interpretation of the improper integral so that when it exists, it is
independent of the choice of limits of integration. Violation of this
property is considered inconsistent with the development of calculus,
more so than trying to preserve some aspect of symmetry in the
integrand. That is the crux of the matter.
Still, the CPV exists and is not without its uses, for example, in
the computation of Hilbert transforms, or other more advanced topics
in real and complex analysis. But it is not generally needed in
elementary calculus, so it is not discussed.
I hope I've answered your question to your satisfaction, long as my
response may be. You've asked an excellent question, one that needs
careful explanation and is not easily dismissed. If I may say so,
it reveals your potential as a mathematician! If you want further
clarification, or have any other questions, feel free to write.
- Doctor Pete, The Math Forum
http://mathforum.org/dr.math/

Date: 09/18/2002 at 11:51:49
From: John Harter
Subject: Thank you (Improper Integral Theory)
Doctor Pete,
Thank you very much for your answer. It was very clear and helpful
(and I much prefer long responses to short ones!). I knew there was
some important reason for defining an improper integral this way, and
I now know that reason!
Thank you again,
John Harter.